What is the integral from 2 to x^2 of cos(t^2+1) ?

The integral from 2 to x^2 of cos(t^2+1) is

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integral from 2 to x^2 of cos(t^2+1)

\int_{2}^{x^{2}} cos (t^{2} + 1) d t

cos (1) (\frac{π}{2} C (\frac{2}{π} x^{2}) - \frac{π}{2} C (\frac{2}{π} \cdot 2)) - sin (1) (\frac{π}{2} § (\frac{2}{π} x^{2}) - \frac{π}{2} § (\frac{2}{π} \cdot 2))

Solution steps

\int_{2}^{x^{2}} cos (t^{2} + 1) d t

Rewrite using trig identities

= \int_{2}^{x^{2}} cos (t^{2}) cos (1) - sin (t^{2}) sin (1) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{x^{2}} cos (t^{2}) cos (1) d t - \int_{2}^{x^{2}} sin (t^{2}) sin (1) d t

\int_{2}^{x^{2}} cos (t^{2}) cos (1) d t = cos (1) (\frac{π}{2} C (\frac{2}{π} x^{2}) - \frac{π}{2} C (\frac{2}{π} \cdot 2))

\int_{2}^{x^{2}} sin (t^{2}) sin (1) d t = sin (1) (\frac{π}{2} § (\frac{2}{π} x^{2}) - \frac{π}{2} § (\frac{2}{π} \cdot 2))

= cos (1) (\frac{π}{2} C (\frac{2}{π} x^{2}) - \frac{π}{2} C (\frac{2}{π} \cdot 2)) - sin (1) (\frac{π}{2} § (\frac{2}{π} x^{2}) - \frac{π}{2} § (\frac{2}{π} \cdot 2))

d er i v a t i v e y = x cos (x)

a re a x = 63 - 7 y^{2}, x = 7 y^{2} - 63

d er i v a t i v e t^{3} - t

\frac{d}{d x} (1 - x^{2} - x arccos (x))

y \frac{d y}{d x} = - 8 cos (π x)

What is the integral from 2 to x^2 of cos(t^2+1) ?
The integral from 2 to x^2 of cos(t^2+1) is